Problem: Evaluate $~~\int\ln(\sqrt{x})\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $=x\ln(x)-x+C$ (Choice B) B $=\dfrac12x\ln(x)-\dfrac12x+C$ (Choice C) C $=x\ln(\sqrt x)-\dfrac23x^\frac32+C$ (Choice D) D $=\dfrac12x\ln(\sqrt{x})-\dfrac13x^\frac32+C$
Explanation: We can first rewrite this problem as $~~\int\ln(x^{1/2})\,dx = \dfrac12\int\ln x\,dx\,$. We will solve this by integrating by parts. We know that $ \int u(x)v\,^\prime(x)dx = u(x)v(x)-\int u\,^\prime(x)v(x)dx\,$. We can rewrite this as $ \int u\ dv = uv-\int v\ du\,$. In this problem we will let $~u = \ln x~$ and $~dv= dx\,$. Then $~du = \dfrac1x\,dx~$ and $~v = x\,$. Integration by parts gives $ \dfrac12\int \ln x\,dx = \dfrac12\Big(x\ln x-\int x\cdot\dfrac1x\,dx\Big)$ $ ~~=\dfrac12x\ln x-\dfrac12x+C\,$.